KAHIBARO
Table of Contents
Meaning and Purpose
In quantum mechanics, a measurement outcome is generally not certain. Even if a particle is described by a perfectly known wave function, measuring its position, momentum, or energy can give different results in repeated experiments. Because of this, we often want a single number that represents the average result of many identical measurements. This number is called the expectation value.
The expectation value is not a guess about one single measurement. Instead, it is the statistical average you would obtain if you prepared the same quantum state many times and measured the same observable each time.
If the observable is represented by an operator $\hat{A}$, then its expectation value in the state $\psi$ is written as
$$
\langle A \rangle = \int \psi^*(x)\,\hat{A}\,\psi(x)\,dx
$$
for a one dimensional continuous system.
Here, $\psi^*(x)$ is the complex conjugate of the wave function, and the integral is taken over all space.
The expectation value of an observable $A$ in the state $\psi$ is
$$
\langle A \rangle = \int \psi^*(x)\,\hat{A}\,\psi(x)\,dx
$$
This formula is valid when the wave function is normalized.
Expectation Value as a Weighted Average
To understand this idea, begin with position. The quantity $|\psi(x)|^2$ gives the probability density for finding the particle at position $x$. So the average position is found by weighting each position by its probability:
$$
\langle x \rangle = \int x\,|\psi(x)|^2\,dx
$$
This is just like an ordinary weighted average in probability.
In the same way, the expectation value of $x^2$ is
$$
\langle x^2 \rangle = \int x^2\,|\psi(x)|^2\,dx
$$
These averages are useful because they describe the center and spread of the probability distribution.
Position Expectation Value
For position, the operator is simply multiplication by $x$. So the expectation value is especially simple:
$$
\hat{x} = x
$$
and therefore
$$
\langle x \rangle = \int \psi^*(x)\,x\,\psi(x)\,dx = \int x\,|\psi(x)|^2\,dx
$$
If the wave function is concentrated more on the right side, then $\langle x \rangle$ will be positive. If it is centered at the origin symmetrically, then $\langle x \rangle$ may be zero.
For example, if $|\psi(x)|^2$ is symmetric about $x=0$, then positive and negative contributions cancel, giving
$$
\langle x \rangle = 0
$$
This does not mean the particle is located at $x=0$. It means the average over many measurements is zero.
Momentum Expectation Value
For momentum in one dimension, the operator is
$$
\hat{p} = -i\hbar \frac{d}{dx}
$$
So the expectation value of momentum is
$$
\langle p \rangle = \int \psi^*(x)\left(-i\hbar \frac{d}{dx}\right)\psi(x)\,dx
$$
This quantity gives the average momentum measured over many identical experiments.
If the wave function represents a state moving mainly to the right, $\langle p \rangle$ is positive. If it is moving mainly to the left, $\langle p \rangle$ is negative.
Important one dimensional expectation values:
$$
\langle x \rangle = \int x\,|\psi(x)|^2\,dx
$$
$$
\langle p \rangle = \int \psi^*(x)\left(-i\hbar \frac{d}{dx}\right)\psi(x)\,dx
$$
Energy Expectation Value
If the Hamiltonian operator is $\hat{H}$, then the expectation value of energy is
$$
\langle E \rangle = \langle H \rangle = \int \psi^*(x)\,\hat{H}\,\psi(x)\,dx
$$
This gives the average energy measured in that quantum state.
If the system is in an energy eigenstate, then every energy measurement gives the same value, and the expectation value equals that eigenvalue exactly. In a more general state, different energy outcomes are possible, and the expectation value is their average.
Discrete Form of Expectation Values
Sometimes an observable has discrete possible values $a_n$, with probabilities $P(a_n)$. In that case, the expectation value is a sum:
$$
\langle A \rangle = \sum_n a_n P(a_n)
$$
This is the direct quantum version of an ordinary statistical mean.
For a state expanded in eigenstates of $\hat{A}$,
$$
\psi = \sum_n c_n \phi_n
$$
the probability of measuring $a_n$ is
$$
P(a_n) = |c_n|^2
$$
so
$$
\langle A \rangle = \sum_n a_n |c_n|^2
$$
Simple Example
Suppose a particle can be found only in two energy states. The possible energies are $E_1$ and $E_2$, and the probabilities are $|c_1|^2$ and $|c_2|^2$, where
$$
|c_1|^2 + |c_2|^2 = 1
$$
Then the expectation value of energy is
$$
\langle E \rangle = E_1 |c_1|^2 + E_2 |c_2|^2
$$
If $|c_1|^2 = 0.7$ and $|c_2|^2 = 0.3$, then the average measured energy lies closer to $E_1$ than to $E_2$.
Expectation Value and Eigenstates
Expectation values become especially simple in an eigenstate. If
$$
\hat{A}\psi = a\psi
$$
then
$$
\langle A \rangle = \int \psi^*\,\hat{A}\,\psi\,dx
= \int \psi^*\,a\psi\,dx
= a \int |\psi|^2 dx
$$
For a normalized wave function,
$$
\int |\psi|^2 dx = 1
$$
so
$$
\langle A \rangle = a
$$
This means that if the system is in an eigenstate of $\hat{A}$, the expectation value equals the corresponding eigenvalue.
If $\psi$ is an eigenstate of $\hat{A}$ with eigenvalue $a$, and $\psi$ is normalized, then
$$
\langle A \rangle = a
$$
In this case, every measurement of $A$ gives the same value $a$.
Expectation Value Is Not Always a Possible Single Result
A very important point is that an expectation value need not be one of the actual measurement outcomes.
For example, if the only possible measured energies are $2 \,\text{eV}$ and $6 \,\text{eV}$, each with probability $1/2$, then
$$
\langle E \rangle = \frac{1}{2}(2) + \frac{1}{2}(6) = 4 \,\text{eV}
$$
But a single measurement never gives $4 \,\text{eV}$. It gives either $2 \,\text{eV}$ or $6 \,\text{eV}$.
So the expectation value is an average over many trials, not necessarily a direct single outcome.
Expectation Values of Powers and Spread
Expectation values are also used to describe how spread out a quantity is. For position, the spread is related to $\langle x^2 \rangle$ and $\langle x \rangle$.
The variance in position is
$$
(\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle^2
$$
and similarly for momentum,
$$
(\Delta p)^2 = \langle p^2 \rangle - \langle p \rangle^2
$$
These quantities measure uncertainty in a statistical sense.
Variance is computed from expectation values:
$$
(\Delta A)^2 = \langle A^2 \rangle - \langle A \rangle^2
$$
This tells us how widely measurement results are distributed around the average.
Summary Table
| Observable | Operator | Expectation value | ||
|---|---|---|---|---|
| Position | $\hat{x} = x$ | $\langle x \rangle = \int x | \psi(x) | ^2 dx$ |
| Momentum | $\hat{p} = -i\hbar \frac{d}{dx}$ | $\langle p \rangle = \int \psi^*(x)\left(-i\hbar \frac{d}{dx}\right)\psi(x)\,dx$ | ||
| Energy | $\hat{H}$ | $\langle H \rangle = \int \psi^*(x)\hat{H}\psi(x)\,dx$ |
Visual Picture
The expectation value of position can be thought of as the balance point of the probability distribution $|\psi(x)|^2$.
Final Idea
Expectation values connect the wave function to measurable average quantities. They are one of the main ways quantum mechanics makes predictions. The wave function gives probabilities, operators represent observables, and the expectation value tells us the average result of many measurements.